Norm of Riemannian curvature tensor on an Einstein manifold under the RIcci flow I'm trying to prove that if $(M, g_0)$ is Einstein and $g(t) = (1-2\lambda t)g_0$ is a solution to its Ricci flow, then $\|\operatorname{Rm}(t)\|^2 = C R(t)^2$, where $R$ is the scalar curvature and $C$ is a constant depending only on $g_0$. This is a claim I've seen in a book but I'm not sure which norm they're using. Assuming it's the one induced by the metric, we have: $\|\operatorname{Rm}\|^2 = g^{ri}g^{sj}g^{pk}g^{q\ell}R_{rspq}R_{ijk\ell}$, but how can we relate this to the Ricci tensor? I'd appreciate any help. This is the discussion in the book: 


 A: Third Edit (the charm??):
I don't know what Chow and Knopf had in mind when they wrote the equation $|\text{Rm}(t)| = CR$, because it's just wrong. But there's a very simple argument to show that the Ricci flow on a positive-scalar-curvature Einstein manifold has a Type I singularity.
Since the evolution of the meetric has the form $g(t) = a(t) g_0$ for some positive function $a(t)$, the inverse metric, $(0,4)$-Riemann curvature tensor, and scalar curvature transform as follows:
$$\begin{align}
\text{Rm}(x,t) &= a(t)\,\text{Rm(x,0)},\\
R(x,t) &= a(t)^{-1} R(x,0),\\
g(x,t)^{-1} & = a(t)^{-1} g_0(x)^{-1}.
\end{align}
$$
(See Theorem 7.30 in my Introduction to Riemannian Manifolds.)
Thus the squared norm of $\text{Rm}(x,t)$ satisfies
$$
\begin{align}
\|\text{Rm}(x,t)\|^2 &= g(x,t)^{-1}*g(x,t)^{-1}*g(x,t)^{-1}*g(x,t)^{-1} * \text{Rm}(x,t)*\text{Rm}(x,t) \\
&=a(t)^{-2}\|\text{Rm(x,0)}\|^2
\end{align}
$$
(where the asterisks represent contractions on various indices). Plugging in $a(t) = C(T-t)$ shows that $\|\text{Rm}(x,t)\|(T-t)$ is constant for each $x$. Since a complete Einstein manifold with positive scalar curvature must be compact by Myers's theorem, this quantity is bounded on $M$.
Here's why $|\text{Rm}(t)| = CR$ (or maybe they meant $CR(t)$) is not true no matter how you interpret it. If you interpret the left-hand side as a pointwise norm, it can't be correct because $R(t)$ is constant for each $t$ while $|\text{Rm}(t)|$ is not. If you interpret $|\text{Rm}(t)|$ to mean the global $L^2$ norm (i.e., the square root of the integral of the squared pointwise norm), then it's not true either because when $g(t) = a(t) g_0$ we have $R(t) = a(t)^{-1}R(0)$ while
$$
\left(\int_M |\text{Rm}(t)|^2dV_{g(t)}\right)^{1/2} = 
a(t)^{-1+n/4}\left(\int_M |\text{Rm(0)}|^2dV_{g_0}\right)^{1/2}.
$$
(See Theorem 7.30 in IRM.)
